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Inessa05 [86]
4 years ago
14

In ABC, and intersect each other at point Q. According to a theorem on medians, Q divides in the ratio 2 : 1. What are the coord

inates of Q?
(5, -3.5)

(4, -5)

(3.33, -3)

(3.67, -4)

Mathematics
1 answer:
Liula [17]4 years ago
6 0
Question b is your answer
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Evaluate the following expressions given x=6 and y=-3
Nana76 [90]

a.

\purple{5 \times 6 - 4 \times ( - 3)}

= 30 + 12

\blue{ =  \boxed{42}}

b.

\purple{ {6}^{2}  +  \frac{6}{ - 3} }

= 36 - 2

\blue{ =  \boxed{34}}

c.

\purple{(6 - 5 {)}^{3}  +  {( - 3)}^{2} }

=  {1}^{3} + 9

= 1 + 9

\blue{ =  \boxed{10}}

<h2>=============================</h2>

Hope it helps

<h2>=============================</h2>
7 0
3 years ago
Homework, 12 points for whoever answers.
adell [148]
The answer to number 1 = 40
5 0
3 years ago
Read 2 more answers
Please match the following 3 terms to their definitions.
yan [13]

Answer:

See attached

Step-by-step explanation:

Refer to picture

8 0
3 years ago
From a deck of five cards numbered 2, 4, 6, 8, and 10, respectively, a card is drawn at random and replaced. this is done three
Serga [27]

Since the sum of the numbers on the three draws is 12, if we want the card numbered 2 to be drawn exactly two times, the third card can only be numbered 8. In fact, 2+2+8 = 12, and there are no other possibilities, unless you consider the various permutations of the terms.

So, we have three favourable cases: we can draw 2,2,8, or 2,8,2, or 8,2,2. This are the only three cases where the card numbered 2 is drawn exactly two times, and the sum of the number on the three draws is 12.

Now, the question is: we have three favourable cases over how many? Well, we have 5 possible outcomes with each draws, and the three draws are identical, because we replace the card we draw every time.

So, we have 5 possible outcomes for the first draw, 5 for the second and 5 for the third. This leads to a total of 5 \times 5 \times 5 = 5^3 = 125 possible triplets.

Once we know the "good" cases and the total number of possible cases, the probability is simply computed as

P = \cfrac{\text{number of favourable cases}}{\text{number of all possible cases}} = \cfrac{3}{125}

3 0
4 years ago
If you multiply 235 by 49, what will be the partial products?
Olenka [21]

Answer:

Step-by-step explanation:

(200+30+5)(40+9)

40*200 = 8000

40*30=   1200

40*5=200

9*200= 1800

9*30=270

9*5=45

8000+1200+200+1800+270+45 =11515

7 0
3 years ago
Read 2 more answers
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